为什么std ::没有比“<"更好? [英] Why is std::less better than "<"?

问题描述

C ++入门，第5部分，14.8.2，将库函数对象与算法配合使用:

C++ primer, 5th, 14.8.2, Using a Library Function Object with the Algorithms:

vector<string *> nameTable;  // vector of pointers
// error: the pointers in nameTable are unrelated, so < is undefined
sort(nameTable.begin(), nameTable.end(),
     [](string *a, string *b) { return a < b; });
// ok: library guarantees that less on pointer types is well defined
sort(nameTable.begin(), nameTable.end(), less<string*>());

然后我检查了std :: less实现:

Then I checked the std::less implementation:

template<typename _Tp>
struct less : public binary_function<_Tp, _Tp, bool>
{
  bool
  operator()(const _Tp& __x, const _Tp& __y) const
  { return __x < __y; }
};

我发现std :: less还使用运算符<做这项工作，为什么<是未定义的，并且库保证对指针类型的定义较少，为什么建议不要使用std:less，为什么要优于<.

I found out that std::less also use operator < to do the work, so why < is undefined and library guarantees that less on pointer types is well defined, why is std:less recommended, and why is std::less better than <.

推荐答案

因为< 并不总是 operator<().只有类才具有运算符功能，因此您的建议不适用于内置类型.

Because < isn't always operator<(). Only classes have operator functions, so your suggestion would not work for the built-in types.

此外，指针上的< 不一定实现严格的弱排序，而 std :: less (通过专门化—发布的内容并不一定 std :: less 的全部"！)必填到:

Furthermore, < on pointers doesn't necessarily implement a strict-weak ordering, whereas std::less (via specialisation — what you posted isn't "all" of std::less!) is required to:

std :: less的特殊化对于任何指针类型都会产生严格的总顺序，即使内置运算符<没有.

A specialization of std::less for any pointer type yields a strict total order, even if the built-in operator< does not.

简而言之: std :: less 适用于支持小于比较的任何事物.

In short: std::less works for anything that supports a less-than comparison.

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